Current mirrors are commonly used for current sources in integrated circuit design. This section covers other current sources that are often seen.

Transcription

1 c Coyright W. Marshall Leach, Jr., Professor, Georgia Institute of Technology, School of Electrical and Comuter Engineering. Current Sources Current mirrors are commonly used for current sources in integrated circuit design. This section covers other current sources that are often seen. FET Current Sources Figure 1(a) shows two FET current sources, one which uses an n-channel deletion mode MOSFET and the other which uses an n-channel JFET. The equivalent -channel sources are shown in Fig. 1(b). Remember that the JFET is a deletion mode device. The analysis for the two sources is the same with the excetion that the transconductance arameter is denoted by K for the MOSFET and by β for the JFET. Figure 1: MOSFET (deletion mode) and JFET current sources. (a) n-channel. (b) -channel. For the n-channel device, the MOSFET drain current and gate-source voltage are given by = K (V GS V TO ) 2 V GS = I S R S The object is to solve for R S for a desired drain current. When the equation for V GS is substituted into the equation for,weobtain ThiscanbesolvedforR S to obtain = K ( R S V TO ) 2 = K ( R S + V TO ) 2 ID /K V TO where V TO < 0. Note that K = K 0 (1 + λv DS ). If V DS is not secified, an often used aroximation is K ' K 0. For the n-channel JFET, the drain current is given by = β (V GS V TO ) 2 It follows that the MOSFET solution for R S canbeusedwiththesubstitutionofβ for K to obtain ID /β V TO 1

2 The outut resistance is a figure of merit for a current source. Ideally, it should be infinite. The outut resistance is the resistance seen looking into the drain of each source. It is given by r out = r id = r 0 1+ R S + R S r s = 1 r s g m r 0 = λ 1 + V DS where g m =2 K for the MOSFET and g m =2 β for the JFET. For the -channel devices, the subscrits for the voltages are reversed, e.g. V GS become V SG and V DS becomes V SD. Examle 1 A deletion mode MOSFET has the arameters K 0 = A/ V 2, λ =10 4 V 1,and V TO = 2V. Calculate the value of R S and r out if the transistor is to be used as a current source with a current =1.5mA. Assume V DS =8V. Solution. K = K 0 (1 + λv DS )= A/ V 2 ID /K V TO =2.47 kω r 0 = λ 1 + V DS =13.87 kω g m =2 K = S r s = 1 g m =566.1 Ω r out = r id = r 0 1+ R S kω r s Resistor R S causes r out to be greater than r 0 by more than a factor of 5. One-BJT Current Source Figure 2 shows nn and n BJT current sources. The object is to select the resistors in the circuit for a desired collector current. Figure 2: BJT current sources. 2

3 For the nn device, the stes can be summarized as follows: (a)chooseavalueforr E. It should be small but large enough for good bias stability. It is common to choose R E so that the voltage across it is some multile n of the base-emitter junction voltage V BE,where tyically 1 n 4. Thelargern, the better the current stability. It follows that I E R E = α R E = nv BE = R E = αnv BE (b)chooseavalueforr 2. It is usually chosen so that the current I 2 is some multile m of I B,where tyically m 9. It follows that I 2 R 2 = mi B R 2 = m β R 2 =(n +1)V BE = R 2 = β (n +1)V BE m If m is too small, the uncertainty in I B can cause errors if β is not known recisely or if β drifts with temerature. (c) Solve for R 1. I 1 R 1 = (m +1)I B R 1 =(m +1) β R 1 = V + V +(n +1)V BE = R 1 = β V + V (n +1)V BE (m +1) (d) Solve for r out. r out = r ic = r 0 + r 0 ekr E 1 αr E r 0 e + R E r 0 = V A + V CE r 0 e = R 1kR 2 + r x For the n device, the subscrits for the voltages are reversed, e.g. V BE become V EB and V CE becomes V EC. Examle 2 A BJT has the arameters β = 100, V A =75V,andr x =40Ω. The transistor is to be used asacurrentsourcewithacurrent =1.5mA, V + =15V,andV = 15 V. Calculate the values of R 1,,R 2,andr out if I 1 =10I B (n = 10) and I E R E =2V BE (m =2). Assume V BE =0.65 V and V CE =8V. + r e Solution. R E = αnv BE = βnv BE (1 + β) = 858 Ω R 1 = β V + V (n +1)V BE (m +1) = 170 kω R 2 = β (n +1)V BE m =13kΩ r 0 = V A + V CE =55.33 kω r 0 e = R 1kR 2 + r x + r e = R 1kR 2 + r x + αv T = Ω r out = r ic = r 0 + r 0 ekr E 1 αr E r 0 e + R E = 380.4kΩ Resistor R E causes rout to be greater than r 0 by almost a factor of 7. 3

4 Two-BJT Current Source Figure 3 shows nn and n two-transistor BJT current sources that are simler to design. The outut current in each is the collector current I O. For the circuit of Fig. 3(a), the following equations can be written V + V = I 1 R 1 + V BE2 + V BE1 + 1 IO I C1 R E2 = V BE1 + 1 α 2 β 1 1 = I 1 I O β 2 For the n circuit, the subscrits for the voltages are reversed, e.g. V BE become V EB. Figure 3: Two-BJT current sources. There is a ositive feedback effect in these circuits which increases the outut resistance seen looking into the collector of Q 1. To see this, consider the small-signal circuit with V + = V =0and the collector of Q 1 driven by a small-signal test current source i t.ifi t is ositive, it causes a current to flow through r 01 from its collector to its emitter. This forces the base voltage of Q 2 to increase. This is amlified by Q 2 to cause its collector voltage to decrease. This decrease is alied to the base of Q 1 to cause its collector voltage to increase. This feedback effect causes the collector voltage of Q 1 to be larger than it would be without the feedback. Because resistance is voltage divided by current, it follows that the resistance seen looking into the collector of Q 1 is increased. The circuit can be designed by secifying the currents I 1 and 1 and the resistor R E2. The current I 1 must be chosen so that it is much larger than the anticiated base current in Q 1. R E2 can be omitted, but it aids in reventing temerature drift of the currents. If it is too large, however, it reduces the gain around the ositive feedback loo, thus reducing the outut resistance. Tyically, R E2 mightbechosentohavea value such that V BE1 I E1 4V BE1.ResistorsR 1 and are then given by V + V (V BE1 + V BE2 ) I 1 I O RE1 β R 1 = 2 R E2 = I 1 V BE1 + 1 I O 1 α 2 β 1 4

5 Examle 3 For V + = 15V and V = 15 V, design the circuit in Fig. 3(a) for an outut current 1 =1.5mA. Solution. If we estimate β = 100, the base current in Q 1 is ma. LetuschooseI 1 =20I B2 =0.3mA. We estimate V BE1 = V BE2 =0.65 V. The current 1 is given by 1 = I 1 I O /β =0.285 ma. The current through is I E1 = 1 /α =0.288 ma. If we choose I E1 = V BE1, it follows that, R 1,andR E2 are given by = 0.65 =2.26 kω ma ma 2.26 kω R 1 = =91.2kΩ 0.3mA R E2 = mA = 867 Ω Current Sources Using an O Am Figure 4 shows two current sources that use an o am as an error amlifier. The emitter current in each transistor is I E = I O α This current divides between R F and R E go cause the voltage at the negative o-am inut to be V N = α R 1 R E + R F + R 1 R E Because the o am forces V N = V I, it follows that I O is given by I O = + R E + R F R 1 VI R E Figure 4: Current sources using an o am. 5